Continuity and Differentiability — Class 12 Maths Solution

Let $y = \log (\cos {e^x})$

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{d}{{dx}}\log (\cos {e^x}) = \cfrac{1}{{\cos {e^x}}}\cfrac{d}{{dx}}(\cos {e^x})$

$= \cfrac{1}{{\cos {e^x}}} \cdot ( - \sin {e^x})\cfrac{d}{{dx}}({e^x}) = - \tan {e^x} \cdot {e^x} = - {e^x}\tan {e^x},$

where ${e^x} \ne (2n + 1)\cfrac{\pi }{2},n \in N$